If the radius of the ${}_{13}^{27}\mathrm{Al}$ nucleus is taken to be $R_{\mathrm{Al}}$, then the radius of ${

Question

If the radius of the ${}_{13}^{27}\mathrm{Al}$ nucleus is taken to be $R_{\mathrm{Al}}$, then the radius of ${}_{53}^{125}\mathrm{Te}$ nucleus is nearly:

Accepted Answer

1. **Formula:** The radius $R$ of a nucleus is related to its mass number $A$ by the relationship $R = R_0 A^{1/3}$, where $R_0$ is a constant .
2. **Identify Mass Numbers:**
   - For Aluminum (${}^{27}_{13}\mathrm{Al}$), the mass number $A_1 = 27$.
   - For Tellurium (${}^{125}_{53}\mathrm{Te}$), the mass number $A_2 = 125$.
3. **Set up Ratio:**
   $$ \frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}} = \frac{R_0 (125)^{1/3}}{R_0 (27)^{1/3}} $$
   $$ \frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}} = \left(\frac{125}{27}\right)^{1/3} $$
4. **Calculate:**
   - Cube root of 125 is 5 ($5^3 = 125$).
   - Cube root of 27 is 3 ($3^3 = 27$).
   $$ \frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}} = \frac{5}{3} $$
   $$ R_{\mathrm{Te}} = \frac{5}{3} R_{\mathrm{Al}} $$

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